List-I (Precipitating reagent and conditions) | List-II (Cation) |
---|---|
(A) \(NH_4Cl + NH_4OH\) | (I) Mn2+ |
(B) \(NH_4OH + Na_2CO_3\) | (II) Pb2+ |
(C) \(NH_4OH + NH_4Cl + H_2S gas\) | (III) Al3+ |
(D) dilute HCl | (IV) Sr2+ |
The matching is based on the chemical properties and specific precipitation conditions for cations:
[A.] NH$_4$Cl + NH$_4$OH precipitates Al(OH)$_3$, hence matches with III. Al$^{3+}$.
[B.] NH$_4$OH + Na$_2$CO$_3$ precipitates SrCO$_3$, hence matches with IV. Sr$^{2+}$.
[C.] NH$_4$OH + NH$_4$Cl + H$_2$S gas precipitates PbS, hence matches with II. Pb$^{2+}$.
[D.] Dilute HCl precipitates MnCl$_2$, hence matches with I. Mn$^{2+}$.
Thus, the correct order is:
\[A -- III, B -- IV, C -- II, D -- I.\]
Match List I with List II:
Choose the correct answer from the options given below:
The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ____. Use: Atomic mass of N (in amu) = 14
The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions is:
List-I | List-II |
---|---|
(P) Passing H2S in the presence of NH4OH | (1) Cu2+ |
(Q) (NH4)2CO3 in the presence of NH4OH | (2) Al3+ |
(R) NH4OH in the presence of NH4Cl | (3) Mn2+ |
(S) Passing H2S in the presence of dilute HCl | (4) Ba2+ (5) Mg2+ |
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: